Given quadratic equation is $\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0$ Here, $\alpha \times \beta$ are roots of equation.


Take square both sides,
$$
\alpha^2+\beta^2+2 \alpha \beta=b^2
$$
Here, $\alpha^2+\beta^2=5$
$$
\begin{aligned}
& 5+2 \alpha \beta=b^2 \\
& 2 \alpha \beta=b^2-5
\end{aligned}
$$

From (i) and take cube both sides, $\alpha^3+\beta^3+3 \alpha \beta(\alpha+\beta)=-b^3$
Here, $\alpha^3+\beta^3=9, \alpha+\beta=-b, \alpha . \beta=\frac{b^2-5}{2}$
$$
\begin{aligned}
& 9+3\left(\frac{b^2-5}{2}\right)(-b)=-b^3 \\
& 18-3 b^3+15 b=-2 b^3 \\
& 18+15 b=b^3 \\
& \Rightarrow b^3-15 b-18=0
\end{aligned}
$$
Now, put $b=-3$ in above equation.
$$
\begin{aligned}
& \Rightarrow(-3)^3+45-18=0 \\
& 45-45=0 \\
& 0=0
\end{aligned}
$$
So, $(b+3)=0$ divides the above equation then, $b=-3$
From the given equation $\mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0$
$$
\Rightarrow \alpha \beta=\mathrm{c}
$$
From (ii),
$$
\alpha \beta=\frac{b^2-5}{2}=c
$$
Put $b=-3$ in above expression,
$$
\begin{aligned}
& \frac{(-3)^2-5}{2}=\mathrm{c} \\
& \mathrm{c}=\frac{9-5}{2}=\frac{4}{2}=2
\end{aligned}
$$
So, $\mathrm{b}+\mathrm{c}=-3+2=-1$.
Therefore, option (b) is correct.